Y2KDREAD

The question is not dumb, i just don't like math.

A box contains two defective Christmas tree lights that have inadvertently mixed with eight nondeffective lights. IF the lights are selected one at a time without replacement and tested until both defective lights are fund, what is the probability that both defective lights will be found after three trials.

I think that you work it out like this to find the answer, but i am not sure:
(2/10)*(2/9)*(1/8)=(1/180)

Thanks

__________________
Signatures are for chumps.

MSD

Exactly three trials, or by the time the third trial takes place?

Y2KDREAD

Exactly three trials, and by the time you finish them you found both of the defective lights

__________________
Signatures are for chumps.

molloby

I think it should be:

(2/10)(2/9)(1/8) + (2/10)(1/9)(1/8) = 1/120

As you can either get the light first or second, but the last one must be deffective.

itch vaccine

first try
you get defective/nondefective/defective

(2/10) (8/9) (1/8) = 1/45

second try
nondefective/defective/defective

(8/10) (2/9) (1/8) = 1/45

probability = 1/45 + 1/45 = 2/45


there couldn't be defective/defective/nondefective because according to you, it must be exactly 3 trials, so why would you want to try again after you get 2 defectives.

however.. if you want to consider defective/defective/nondefective
then it should be
2/45 + (2/10) (1/9) (8/8) = 1/15

Hope this helps.

molloby

Well darn, I was wrong.

Thermopyle

Good work Itch...

__________________
Live life like you're gonna die, beacause you're gonna!

- William Shatner.

wgheow

fuiyoh..good